The equation frac{x^2}{12 - lambda} + frac{y^ | vedclass

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(C) The given equation is $\frac{x^2}{12 - \lambda} + \frac{y^2}{8 - \lambda} = 1$.Let $a^2 = 12 - \lambda$ and $b^2 = 8 - \lambda$.Case $1$: If $8 <

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If $8 < \lambda < 12$,then it is a hyperbola.

Vedclass · Answer

(C) The given equation is $\frac{x^2}{12 - \lambda} + \frac{y^2}{8 - \lambda} = 1$.Let $a^2 = 12 - \lambda$ and $b^2 = 8 - \lambda$.Case $1$: If $8  0$ and $8 - \lambda Let $12 - \lambda = a^2$ and $8 - \lambda = -b^2$ (where $b^2 > 0$).The equation becomes $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,which is the standard equation of a hyperbola.Therefore,if $8 < \lambda < 12$,the equation represents a hyperbola.

The equation $\frac{x^2}{12 - \lambda} + \frac{y^2}{8 - \lambda} = 1$ represents:

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